Explanation
We have,
$$ \sin \left( {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x \right)=1 $$
$$ {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x={{\sin }^{-1}}1 $$
$$ {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x=\dfrac{\pi }{2} $$
$$ {{\sin }^{-1}}\dfrac{1}{5}=\dfrac{\pi }{2}-{{\cos }^{-1}}x $$
$$ {{\sin }^{-1}}\dfrac{1}{5}={{\sin }^{-1}}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \left( {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \right) $$
$$ x=\dfrac{1}{5} $$
Hence, this is the answer.
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